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Step2247 [10]
3 years ago
11

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

the curve is 7,500 Newtons. What speed limit should be posted for the curve so that cars can negotiate it safely
Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

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3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

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mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

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Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

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Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

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4 0
3 years ago
The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r
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Answer:

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3 years ago
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